Step 1: Geometry:
On one face of the fcc cube, there is an atom at each corner and one in the centre. The diagonal of the face passes through three atoms touching each other: corner, centre, corner.
Step 2: Length along the diagonal:
That equals $r + 2r + r = 4r$. By Pythagoras, the face diagonal of a square of side $a$ is $a\sqrt{2}$.
Step 3: Solve:
$4r = 360 \times 1.4142 = 509.1$ pm, so $r = 509.1/4 = 127.3$ pm.
Step 4: Check with options:
Only $127.26$ pm matches within rounding.
Final Answer:
The atomic radius is 127.26 pm, option (B).
\[ \boxed{127.26 \text{ pm}} \]