The electrons identified by quantum numbers $n$ and $l$:
(a) $n=4,l=1$ (b) $n=4,l=0$ (c) $n=3,l=2$ (d) $n=3,l=1$
Can be placed in order of increasing energy as:

The correct option is A (d) < (b) < (c) < (a)
$l=0\to s$
$l=1\to p$
$l=2\to d$
$l=3\to f$

Now, $\left(a\right)\to 4p,\left(b\right)\to 4s,\left(c\right)\to 3d,\left(d\right)\to 3s$

According to aufbau principle, increasing order is $3s<4s<3d<4p$ i.e., $d<b<c<a$

Hence,B is correct option.