Write the given relation

$F=ax^2+b t^{1/2}$

Since both terms are added, each term must have the dimensions of force.

$[F]=[MLT^{-2}]$

Find dimensions of $a$

From $ax^2 ext{ has dimensions of force}$

Since $[x]=[L],$ we get $[a][L^2]=[MLT^{-2}]$

Therefore, $[a]=[ML^{-1}T^{-2}]$

Find dimensions of $b$

From $b t^{1/2} ext{ has dimensions of force}$

Since $[t^{1/2}]=[T^{1/2}],$ we get $[b][T^{1/2}]=[MLT^{-2}]$

Therefore, $[b]=[MLT^{-5/2}]$

Find dimensions of $b^2/a$

First, $[b^2]=[M^2L^2T^{-5}]$

Now divide by $[a]$: $\left[\frac{b^2}{a}\right]=\frac{[M^2L^2T^{-5}]}{[ML^{-1}T^{-2}]}$

$\left[\frac{b^2}{a}\right]=[ML^3T^{-3}]$

Match with options

$[ML^3T^{-3}]$ corresponds to Option B.

Comparison with stored answer

Stored correct answer: B

Derived answer: B

Hence, they agree.