Assume we are talking about a standard normal distribution with zero mean and unit variance. For the observer to be able to answer the question "is this sample from a standard normal distribution?" with a high probability of correctness, he needs to know the distribution of distributions from which the sample may have been generated. The probability that the observer will guess "yes" is maximized when the sample is generated from a standard normal distribution, assuming that is possible. So according to my interpretation, you should use the values generated by Box-Muller in step 1 without inspecting them.
First will come the pretty complete theory. Then we look at your particular situation.
Theory: We need an expression for the variance of $X$. The variance is $E(X^2)-(E(X))^2$. For $E(X^2)$, we need to calculate
$$\int_a^b \frac{x^2}{b-a}\,dx,$$
which is $\frac{b^3-a^3}{3(b-a)}$. This simplifies to $\frac{b^2+ab+a^2}{3}$.
I imagine that you know that $E(X)=\frac{b+a}{2}$. One can do this by integration, but it is clear by symmetry that the mean is halfway between $a$ and $b$.
So we know that the variance is $\frac{b^2+ab+a^2}{3}-\frac{(b+a)^2}{4}$. Bring to a common denominator, simplify. We get that
$$\text{Var}(X)=\frac{(b-a)^2}{12} \tag{$\ast$}.$$
More simply, you can search under uniform distribution, say on Wikipedia. They will have the expression $(\ast)$ for the variance of $X$.
Your problem: We know that $\frac{b+a}{2}=3.5$. We also know that the standard deviation of $X$ is $1.3$, so the variance is $(1.3)^2=1.69$.
So, by $(\ast)$, $\frac{(b-a)^2}{12}=1.69$, and therefore $b-a=\sqrt{(12)(1.69)}\approx 4.5033$. We also know that $b+a=(2)(3.5)=7$. Now that we know $b-a$ and $b+a$, it is easy to find $a$ and $b$.
For your simulation, presumably you are starting from a random number generator that generates numbers that are more or less uniformly distributed on $[0,1)$. If $U$ represents the output of such a generator, we simulate $X$ by using $a+(b-a)U$. And we now know $a$ and $b$.
Added If you want a general formula instead of a procedure, let $\mu=\frac{a+b}{2}$ be the mean, and $\sigma=\frac{b-a}{\sqrt{12}}=\frac{b-a}{2\sqrt{3}}$ be the standard deviation. Then $\frac{b-a}{2}=\sqrt{3}\,\sigma$.
We get $a=\frac{b+a}{2}-\frac{b-a}{2}=\mu-\sqrt{3}\,\sigma$. It follows that we can take
$$X=\mu-\sqrt{3}\,\sigma + (2\sqrt{3}\,\sigma) U=\mu+(\sqrt{3}\,\sigma)(2U-1).$$
Best Answer
What copperhat is hinting at is the following algorithm:
(sorry, would be a mess as a comment).
In general, if you have a continuous distribution with cummulative distribution $c(x)$, to generate the respective numbers get $u$ as above, $c^{-1}(u)$ will have the required distribution. Can do the same with discrete distributions, essentially searching where the $u$ lies in the cummulative distribution.
An extensive treatment of generating random numbers (including non-uniform ones) is in Knuth's "Seminumerical Algorithms" (volume 2 of "The Art of Computer Programming"). Warning, heavy math involved.