The answer is d) 6.
Say radius of the rupee is $R$ and another coin ($C_1$) is kept touching the center coin ($C_0$).
Compute the angle subtended at the center of $C_0$ by $C_1$:
- Join the centers by a line of length $2R$.
- Draw a tangent to $C_1$ from center of $C_0$, touching $C_1$ at $T$
- Now draw the radius in $C_1$ from its center to $T$.
In the resulting right angled triangle, the angle $\angle center_1center_0T$ is clearly $30^{\circ}$ (simple trirgonometry, $\sin {30^{\circ}} = \frac{R}{2R}$), thus the angle subtended is $60^{\circ}$, which allows 6 coins to be placed.
Interestingly, this is also follows from the intuitive answer to the tightest packing possible for circles: hexagonal
