Dimensions of universal gravitational constant $$(G)$$ in terms of Planck's constant $$(h)$$, distance $$(L)$$, mass $$(M)$$ and time $$(T)$$ are :

Under Newton’s law, the gravitational force between two point masses is
$$F=\dfrac{G\,m_1 m_2}{r^{2}}$$

Writing the dimensions:
$$[F]=MLT^{-2},\qquad [m_1]=[m_2]=M,\qquad[r]=L$$

Hence
$$MLT^{-2}=G\,M\;M\;L^{-2}$$
which gives the dimensional formula of the universal gravitational constant as
$$[G]=M^{-1}L^{3}T^{-2}$$

Planck’s constant has the well-known dimensions of action (energy × time):
$$[h]=ML^{2}T^{-1}$$

We want to rewrite $$[G]$$ in the form $$h^{a}L^{b}M^{c}T^{d}$$. Assume
$$[G]=[h]^{a}L^{b}M^{c}T^{d}$$

Substituting $$[h]=ML^{2}T^{-1}$$:
$$M^{-1}L^{3}T^{-2}=(M^{1}L^{2}T^{-1})^{a}L^{b}M^{c}T^{d}$$ $$\Rightarrow M^{-1}L^{3}T^{-2}=M^{a+c}\;L^{2a+b}\;T^{-a+d}$$

Equating the exponents of each fundamental dimension:

For mass $$M$$ : $$a+c=-1$$

For length $$L$$ : $$2a+b=3$$

For time $$T$$ : $$-a+d=-2$$

A very simple choice that satisfies all three equations is $$a=1,\; b=1,\; c=-2,\; d=-1$$

Therefore
$$[G]=[h]^{1}L^{1}M^{-2}T^{-1}=h\,T^{-1}L\,M^{-2}$$

That matches Option B.

Answer: Option B which is: $$[hT^{-1}LM^{-2}]$$