Here,

We know that acceleration due to gravity for any planet is given by:

\(\rm g\)=$\frac{GM}{r^2}$

Where,

\(\rm g\)= acceleration due to gravity

\(\rm G\)= Universal gravitational constant

\(\rm M\)= mass of the planet

\(\rm r\)= radius of the planet

Similarly, the acceleration due to gravity on earth is given by:

\(\rm g_{e}\)=$\frac{GM_{e}}{(r_{e})^2}$----(i)

Where,

\(\rm g_{e}\)= acceleration due to gravity on earth

\(\rm G\)= Universal gravitational constant

\(\rm M_{e}\)= mass of earth

\(\rm r_{e}\)= radius of earth

According to question,

If the earth is squeezed to the size of the moon, what will be the effect on its acceleration due to gravity?

When earth is squeezed to the size of moon its mass remains the same.

But since the moon's radius is $\frac{37}{10}$ times smaller than the Earth's radius, so the new radius of the squeezed Earth would be:

\(\rm r_{n}\) = $\frac{r_e}{\frac{37}{10}}$

\(\rm r_{n}\) = $\frac{10}{37}$\(\rm .r_{e}\)

Hence, Now the new acceleration due to gravity would be:

\(\rm g_{n}\)=$\frac{GM_{e}}{(r_{n})^2}$

Substituting \(\rm r_{n}\) = $\frac{10}{37}$\(\rm .r_{e}\) 

\(\rm g_{n}\)=$\frac{GM_{e}}{(\frac{10}{37}.r_{e}  )^2}$   

\(\rm g_{n}\)=$\frac{GM_{e}}{\frac{100}{1369}.(r_{e}  )^2}$   

\(\rm g_{n}\)=$\frac{1369}{100}$ \(\rm . \)$\frac{GM_{e}}{(r_{e})^2}$

\(\rm g_{n}\)=$\frac{1369}{100}$ \(\rm . \) \(\rm g_{e}\)         

{ From eqn(i)}                     

\(\rm g_{n}\)=\(\rm 13.69\) \(\rm . \) \(\rm g_{e}\)

Conclusion:

If the Earth were squeezed to the size of the Moon, its acceleration due to gravity would increase by a factor of approximately \(\rm 13.69\) . Given that the acceleration due to gravity on Earth is about \(\rm 9.8 ms^{-2}\), the new acceleration due to gravity would be:

\(\rm g_{n}\)= \(\rm 13.69\) \(\rm . \) \(\rm 9.8 ms^{-2}\)

\(\rm g_{n}\)=\(\rm 134.16 ms^{-2}\)

Hence , the gravity on the squeezed Earth would be much stronger.

It would be about \(\rm 134.16 ms^{-2}\).