Step 1: Recall how a bright fringe's position on the screen relates to path difference.
A bright fringe of order $n$ forms where the two waves from the slits arrive with a path difference of $n\lambda$, and this fringe sits a definite distance from the centre of the screen.
Step 2: Find the position of the fourth bright fringe.
\[ y_n = \frac{n\lambda D}{d} \] with $d = 0.1$ mm, $D = 4$ m, $n = 4$, $\lambda = 620$ nm: \[ y_4 = \frac{4 \times 620 \times 10^{-9} \times 4}{0.1 \times 10^{-3}} = \frac{9.92 \times 10^{-6}}{10^{-4}} = 9.92 \times 10^{-2}\ \text{m} \]
Step 3: Work the path difference back from this position, using the small-angle geometry of the setup.
\[ \Delta x = \frac{y_4 d}{D} = \frac{9.92 \times 10^{-2} \times 0.1 \times 10^{-3}}{4} = \frac{9.92 \times 10^{-6}}{4} = 2.48 \times 10^{-6}\ \text{m} \]
Step 4: Convert to convenient units and check against the direct formula.
\[ 2.48 \times 10^{-6}\ \text{m} = 2.48\ \mu\text{m} \] This matches $n\lambda = 4 \times 620\ \text{nm} = 2480\ \text{nm} = 2.48\ \mu\text{m}$ exactly, so the geometric route agrees with the order-wavelength relation.
Step 5: Why the other options fail.
$1.24\ \mu\text{m}$ is the path difference for order $2$, not $4$. $155$ nm is a quarter of $620$ nm and matches no bright fringe condition. $0$ only holds at the central maximum, order zero.
Final Answer:
\[ \boxed{2.48\ \mu\text{m}} \]