how will you solve it?

The question should be phrased like this :

"Let $s, t$ be the largest positive integers such that... "

$400!$ has $7^s$ in it's prime representation, where

$s=\left \lfloor \dfrac {400}{7}\right \rfloor +\left \lfloor \dfrac {400}{49}\right \rfloor +\left \lfloor \dfrac {400}{343}\right \rfloor =\boxed {66}$

$((3!)!)!=720!$

$720!$ has $3^t$ in it's prime representation, where

$t=\left \lfloor \dfrac {720}{3}\right \rfloor +\left \lfloor \dfrac {720}{9}\right \rfloor +\left \lfloor \dfrac {720}{27}\right \rfloor +\left \lfloor \dfrac {720}{81}\right \rfloor +\left \lfloor \dfrac {720}{243}\right \rfloor =\boxed {356}$

Therefore $s+t=66+356=\boxed {422}$ .

$S=\left \lfloor \cfrac {400}{7}\right \rfloor +\left \lfloor \cfrac {400}{7^2}\right \rfloor +\left \lfloor \cfrac {400}{7^3}\right \rfloor =\boxed {66}$

$((3!)!)!=720!$

$T=\left \lfloor \cfrac {720}{3}\right \rfloor +\left \lfloor \cfrac {720}{3^2}\right \rfloor +\left \lfloor \cfrac {720}{3^3}\right \rfloor +\left \lfloor \cfrac {720}{3^4}\right \rfloor +\left \lfloor \cfrac {720}{3^5}\right \rfloor =\boxed {356}$

So $S+T=66+356=\boxed {422}$