Your model and its estimates posit that
$$\sqrt{Y} = 2.1014 D - 3.0147 + \varepsilon$$
where $D$ is Dose.Back (or its logarithm) and $\varepsilon$ is a random variable of zero expectation whose standard deviation is approximately $16.28.$ Squaring both sides gives
$$Y = (2.1014 D - 3.0147 + \varepsilon)^2.$$
Adding $0.01$ to $D$ yields the value
$$(2.1014 (D + 0.01) - 3.0147 + \varepsilon')^2.$$
The difference is
$$2(2.1014 D - 3.0147 + \varepsilon)(\varepsilon' - \varepsilon + (0.01)(2.1014)) + (\varepsilon' - \varepsilon + (0.01)(2.1014))^2.$$
This expression, as well as its expectation, are complicated. Let us therefore focus on the simpler question of how the expectation of $Y$ varies with $D$. Note that
$$\eqalign{
\mathbb{E}(Y) &= \mathbb{E}\left(2.1014 D - 3.0147 + \varepsilon\right)^2 \\
&= (2.1014D - 3.0147)^2 + 2(2.1014D - 3.0147) \mathbb{E}(\varepsilon) + \mathbb{E}(\varepsilon^2) \\
&=(2.1014D - 3.0147)^2 + 0 + (16.28)^2.
}$$
(This result is of considerable interest in its own right because it reveals the role played by the mean squared error in interpreting the relationship between $D$ and $Y$.)
When $0.01$ is added to $D$ the value of $\mathbb{E}(Y)$ increases by
$$2(2.1014)(2.1014D - 3.0147)(0.01) + 2.1014(0.01)^2.$$
The last term $2.1014(0.01)^2 \approx 0.0002$ is so small compared to the squared errors (with their typical value of $16.28$) that we may neglect it. In this case, to a good approximation, this fitted model associates an (additive) increase in $D$ of $0.01$ with an increase in $Y$ of
$$2(2.1014)(2.1014D - 3.0147)(0.01) = 0.0883176 D - 0.126.$$
When $D$ is the natural logarithm of some quantity $d$, a 1% multiplicative increase in $d$ causes a value of approximately $0.01$ to be added to $D$, because
$$\log(1.01 d) = \log(1.01) + \log(d) = \left(0.01 - (0.01)^2/2 + \cdots\right) + D \approx 0.01 + D.$$
If you used a logarithm to another base $b$, entailing $D = \log_b(d) = \log(d)/\log(b),$ then a 1% multiplicative increase in $d$ causes a value of approximately $(0.01)/\log(b)$ to be added to $D$, so everywhere "$0.01$" occurs in the preceding formulas you must use $(0.01/\log(b))$ instead.
You presumably log-transformed your data for ANOVA because residuals weren't normally distributed and/or they depended on the magnitudes of the data values. So, for the same reasons, further statistical tests (like pairwise comparisons) should also be performed on the transformed data.
When you write up your results you might back-transform to the original scale to make results easier for a reader who expects to see values in that scale, but note that confidence intervals on the original scale will no longer be symmetric about the mean values.
Best Answer
If $y = \arcsin(\sqrt{p})$ then $p=(\sin(y))^2$.
To convert a proportion to a percentage, multiply by 100.
Note that your original percentages have to be transformed to proportions before taking the arcsin-square-root.