This is a transcendental number, in fact one of the best known ones, it is $6+$ Champernowne's number.
Kurt Mahler was first to show that the number is transcendental, a proof can be found on his "Lectures on Diophantine approximations", available through Project Euclid. The argument (as typical in this area) consists in analyzing the rate at which rational numbers can approximate the constant (see the section on "Approximation by rational numbers: Liouville to Roth" in the Wikipedia entry for Transcendence theory).
An excellent book to learn about proofs of transcendence is "Making transcendence transparent: an intuitive approach to classical transcendental number theory", by Edward Burger and Robert Tubbs.
This is not an answer, just a list of papers concerned with this constant; it's too long for a comment. Going by the reviews of the papers, they deal with efficient methods for calculating the constant, and not (directly) with questions of irrationality and/or transcendence. The first paper on the list is too new to have been reviewed at this point.
MR3349435
Lu, Dawei; Song, Zexi;
Some new continued fraction estimates of the Somos' quadratic recurrence constant.
J. Number Theory 155 (2015), 36–45.
MR3019753
Chen, Chao-Ping
New asymptotic expansions related to Somos' quadratic recurrence constant.
C. R. Math. Acad. Sci. Paris 351 (2013), no. 1-2, 9–12.
MR2825112 (2012h:11181)
Hirschhorn, Michael D.
A note on Somos' quadratic recurrence constant.
J. Number Theory 131 (2011), no. 11, 2061–2063.
MR2809034 (2012e:05038)
Nemes, Gergő
On the coefficients of an asymptotic expansion related to Somos' quadratic recurrence constant.
Appl. Anal. Discrete Math. 5 (2011), no. 1, 60–66.
MR2684487 (2011i:11179)
Mortici, Cristinel
Estimating the Somos' quadratic recurrence constant.
J. Number Theory 130 (2010), no. 12, 2650–2657.
MR2319662 (2008f:40013)
Sondow, Jonathan; Hadjicostas, Petros
The generalized-Euler-constant function $\gamma(z)$ and a generalization of Somos's quadratic recurrence constant.
J. Math. Anal. Appl. 332 (2007), no. 1, 292–314.
MR2262724 (2008b:11081)
Hessami Pilehrood, Khodabakhsh; Hessami Pilehrood, Tatiana
Arithmetical properties of some series with logarithmic coefficients.
Math. Z. 255 (2007), no. 1, 117–131.
Best Answer
The number $0.1010010001000010000010000001\ldots$ is transcendental.
Consider following three Jacobi theta series defined by $$\begin{align} \theta_2(q) &= 2q^{1/4}\sum_{n\ge 0} q^{n(n+1)} = 2q^{1/4}\prod_{n=1}^\infty (1-q^{4n})(1 + q^{2n})\\ \theta_3(q) &= \sum_{n\in\mathbb{Z}} q^{n^2} = \prod_{n=1}^\infty (1-q^{2n})(1+ q^{2n-1})^2\\ \theta_4(q) &= \theta_3(-q) = \sum_{n\in\mathbb{Z}} (-1)^n q^{n^2} = \prod_{n=1}^\infty (1-q^{2n})(1- q^{2n-1})^2\\ \end{align} $$ and for any $m \in \mathbb{Z}_{+}$, $k \in \{ 2, 3, 4 \}$, use $\displaystyle D^m\theta_k(q)$ as a shorthand for $\displaystyle \left( q\frac{d}{dq} \right)^m \theta_k(q)$.
Based on Corollary 52 of a survey article Elliptic functions and Transcendence by M. Waldschmidt in 2006,
We know for any non-zero algebraic $q$ with $|q| < 1$, the three $\theta_k(q)$, in particular $\theta_2(q)$ is transcendental. Since
$$\sum_{n=1}^\infty 10^{-n(n+1)/2} = \frac{\sqrt[8]{10}}{2} \theta_2\left(\frac{1}{\sqrt{10}}\right) - 1$$
and using the fact $\frac{1}{\sqrt{10}}$ and $\frac{\sqrt[8]{10}}{2}$ are both algebraic, we find the number at hand is transcendental.