(d) $\dfrac{3}{4}\pi d^2$
Since radius $r = \dfrac{d}{2}$, TSA of solid hemisphere $= 3\pi r^2 = 3\pi \left(\dfrac{d}{2}\right)^2 = \dfrac{3}{4}\pi d^2$.
TSA of a solid hemisphere = curved surface area + circular base = $2\pi r^2 + \pi r^2 = 3\pi r^2$. Substituting $r = d/2$ gives $\frac{3}{4}\pi d^2$. Students often forget to include the flat circular base, which leads to the wrong option (b).