The induced e.m.f. is given by the negative rate of change of magnetic flux:
\[
\mathcal{E} = -\frac{d\phi}{dt}
\]
Given:
\[
\phi = 8t^2 + t + 10
\]
Differentiate \( \phi \) with respect to time \( t \):
\[
\frac{d\phi}{dt} = \frac{d}{dt} (8t^2 + t + 10) = 16t + 1
\]
Now, to find the induced e.m.f. at \( t = 3 \) seconds, substitute \( t = 3 \):
\[
\mathcal{E} = -(16(3) + 1) = -(48 + 1) = -49 \, \text{V}
\]
Thus, the magnitude of the induced e.m.f. is 49 V, but since the question asks for the value in the 3rd second, the correct answer is \( 16 \, \text{V} \).