1. Gravitational field of a uniform solid sphere

For a solid sphere of radius $R$ and mass $M$:

• Inside the sphere $(r<R)$, $g(r)=\frac{GM}{R^3}r$ so the field increases linearly from the center to the surface.

• At the surface $(r=R)$, $g_{\text{surface}}=\frac{GM}{R^2}$

• Outside the sphere $(r>R)$, $g(r)=\frac{GM}{r^2}$

Thus, in the graph of $g$ vs. $r$, the maximum value occurs at the surface.

2. Use the graph information

Since the statement says that the gravitational fields due to sphere (1) and sphere (2) are shown, the relevant comparison is at their maxima, i.e. at their surfaces.

From the graph, the maximum field of sphere (1) is equal to the maximum field of sphere (2).

So, $\frac{GM_1}{R_1^2}=\frac{GM_2}{R_2^2}$

Cancel $G$: $\frac{M_1}{R_1^2}=\frac{M_2}{R_2^2}$

Hence, $\frac{M_1}{M_2}=\frac{R_1^2}{R_2^2}$

Given: $R_1=1\text{ m},\qquad R_2=2\text{ m}$

Therefore, $\frac{M_1}{M_2}=\frac{1^2}{2^2}=\frac{1}{4}$

3. Compare with options

The value obtained is $\boxed{\frac{M_1}{M_2}=\frac{1}{4}}$

But $\frac{1}{4}$ is not present in the options.

So the stored answer $\frac{1}{6}$ cannot be correct for the standard gravitational-field graph interpretation.

4. Conclusion

Using the standard formula for gravitational field of solid spheres and equating the surface values indicated by the graph, we get: $\boxed{\frac{M_1}{M_2}=\frac{1}{4}}$

Hence I disagree with the stored answer.