Step 1: Picture where atoms touch in a BCC cell.
In a body centred cubic lattice the corner atoms never touch each other along the edge. Instead, the central atom touches the two corner atoms along the body diagonal of the cube. That single line of contact is the key to the whole problem.
Step 2: Write the length of the body diagonal.
For a cube of edge $a$, the body diagonal has length $\sqrt{3}\,a$ (from the three dimensional Pythagoras result).
Step 3: Count radii along that diagonal.
Travelling along the body diagonal we cross half a corner atom, then the whole central atom (two radii), then half the opposite corner atom. That is $r + 2r + r = 4r$.
Step 4: Form the working equation.
Setting the geometric length equal to the atomic count gives $\sqrt{3}\,a = 4r$, so $r = \dfrac{\sqrt{3}}{4}\,a$.
Step 5: Plug in the numbers.
Here $a = 352$ pm. First divide by 4: $\dfrac{352}{4} = 88$. Then multiply by $\sqrt{3} \approx 1.732$: $r = 88 \times 1.732$.
Step 6: Finish the arithmetic.
$88 \times 1.732 = 152.4$ pm (since $88 \times 1.7 = 149.6$ and $88 \times 0.032 = 2.8$, adding to $152.4$). This matches option (3).
\[ \boxed{r \approx 152.4 \text{ pm}} \]