The orbital having two radial as well as two angular nodes is:

For an orbital with quantum numbers $$n$$ and $$l$$, the number of radial (nodal surfaces) nodes is $$(n - l - 1)$$ and the number of angular nodes is $$l$$.

We need an orbital with 2 radial nodes and 2 angular nodes. Since angular nodes $$= l$$, we need $$l = 2$$, which corresponds to a $$d$$ orbital.

For radial nodes: $$n - l - 1 = 2$$, which gives $$n - 2 - 1 = 2$$, so $$n = 5$$.

Let us verify with each option: For $$5d$$: $$l = 2$$, angular nodes $$= 2$$, radial nodes $$= 5 - 2 - 1 = 2$$. This satisfies both conditions. For $$4d$$: radial nodes $$= 4 - 2 - 1 = 1$$ (does not match). For $$4f$$: $$l = 3$$, angular nodes $$= 3$$ (does not match). For $$3p$$: $$l = 1$$, angular nodes $$= 1$$ (does not match).

Therefore, the orbital with two radial and two angular nodes is $$\mathbf{5d}$$.