Step 1: Rolling acceleration formula.
A body rolling down without slipping has $a = \dfrac{g\sin\theta}{1 + \frac{I}{mR^{2}}}$. The factor $\frac{I}{mR^{2}}$ depends on the shape.
Step 2: Use the disc to find the angle.
For a disc $\frac{I}{mR^{2}} = \tfrac{1}{2}$, so $a_{disc} = \tfrac{2}{3}g\sin\theta$. We are told $a_{disc} = \tfrac{1}{3}g$.
Step 3: Solve for $\sin\theta$.
\[ \tfrac{2}{3}g\sin\theta = \tfrac{1}{3}g \quad\Rightarrow\quad \sin\theta = 0.5 \]
Step 4: Acceleration of the solid sphere.
For a sphere $\frac{I}{mR^{2}} = \tfrac{2}{5}$, so $a_{sph} = \tfrac{5}{7}g\sin\theta$. \[ a_{sph} = \tfrac{5}{7}(10)(0.5) = \frac{25}{7} \text{ m s}^{-2} \]
Step 5: Use $v^{2} = 2as$ down the slope.
The length is $s = 1.75$ m and it starts from rest. \[ v^{2} = 2 \times \frac{25}{7} \times 1.75 = \frac{50}{7}\times\frac{7}{4} = 12.5 \]
Step 6: Take the square root.
\[ v = \sqrt{12.5} \approx 3.5 \text{ m s}^{-1} \]So the sphere reaches the bottom at about $3.5$ m s$^{-1}$, which is option 3.
\[ \boxed{v \approx 3.5 \text{ m s}^{-1}} \]