Explanation

(d) The moment of inertia of the disc about an axis parallel to its plane is
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$I_{t} = I_{d} + MR^{2}$
$\Rightarrow$ $I = \frac{1}{4}MR^{2} + MR^{2}$
$= \frac{5}{4}MR^{2}$
or $MR^{2} = \frac{4I}{5}$
Now, moment of inertia about a tangent perpendicular to its plane is
$I^{''} = \frac{3}{2}MR^{2} = \frac{3}{2} \times \frac{4}{5}I = \frac{6}{5}I$