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$\begin{aligned}\frac{dV}{dt}&=a\\V&=at+{V}_{0}\\\frac{4}{3}\pi{r}^3&=at+{V}_{0}\text{ ...1}\\4\pi{r}^2\frac{dr}{dt}&=a\\\frac{dr}{dt}&=\frac{a}{4\pi{r}^2}\\\epsilon&=B\frac{dA}{dt}\\&=2\pi rB\frac{dr}{dt}\\&=2\pi rB\frac{a}{4\pi{r}^2}\\&=\frac{aB}{2r}\\r&={\left(\frac{243}{16}\right)}^{\frac{1}{3}}\text{(from 1)}\\\epsilon&=\frac{2\times 1.5}{2\times {\left(\frac{243}{16}\right)}^{\frac{1}{3}}}\\&\huge\color{#3D99F6}{=\boxed{0.605 V}}\end{aligned}$

Note that $V = \frac{4}{3} \pi r^3 \\ \implies \pi \left ( \frac{3}{4 \pi} \right )^{\frac{2}{3}} V^\frac{2}{3} = \pi r^2 = A$

Now, $\Phi = B A$

And, $\epsilon = - \frac{\partial {\Phi}}{\partial t} \\ = - B \frac{\partial }{\partial t} \left ( \pi \left ( \frac{3}{4 \pi} \right )^{\frac{2}{3}} V^\frac{2}{3}\right )$

Furthermore, we notice that $V(t) = 20 \pi + a t$

Plugging in the above, and doing the appropriate differentiation, we get:

$\epsilon = -\frac{3^{2/3} \sqrt[3]{\frac{\pi }{2}} a}{2 \sqrt[3]{a t+20 \pi }}$

The value of which is equal to $-\frac{\sqrt[3]{2}}{3^{2/3}}$ at $t = \pi / 8$ and $a = 2$