1. Use the formulas for nodes

For an orbital with principal quantum number $n$ and azimuthal quantum number $l$:

• Angular nodes $= l$
• Radial nodes $= n - l - 1$

We need an orbital having:

• $2$ angular nodes
• $2$ radial nodes

2. Find the required quantum numbers

Since angular nodes $= l = 2$, the orbital must be a $d$-orbital.

Now radial nodes should also be $2$:

$n - l - 1 = 2$

Substitute $l=2$:

$n - 2 - 1 = 2$ $n - 3 = 2$ $n = 5$

So the orbital is $5d$.

3. Check each option

A: $3p$

For $p$ orbital, $l=1$.

• Angular nodes $=1$
• Radial nodes $= 3-1-1=1$

So $3p$ has 1 radial and 1 angular node. Not correct.

B: $5d$

For $d$ orbital, $l=2$.

• Angular nodes $=2$
• Radial nodes $= 5-2-1=2$

So $5d$ has 2 radial and 2 angular nodes. Correct.

C: $4d$

For $d$ orbital, $l=2$.

• Angular nodes $=2$
• Radial nodes $= 4-2-1=1$

So $4d$ has 1 radial and 2 angular nodes. Not correct.

D: $4f$

For $f$ orbital, $l=3$.

• Angular nodes $=3$
• Radial nodes $= 4-3-1=0$

So $4f$ has 0 radial and 3 angular nodes. Not correct.

4. Final answer

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

$\boxed{5d}$

So, Option B is correct.