1. Motional emf induced in the rod

A rod of length $l = 1\,\text{m}$ moves with speed $v = 4\,\text{m s}^{-1}$ perpendicular to a magnetic field $B = 0.15\,\text{T}$.

The motional emf is

$\mathcal{E} = Blv$

So,

$\mathcal{E} = 0.15 \times 1 \times 4 = 0.6\,\text{V}$

2. Current induced in the circuit

Given total resistance of the circuit,

$R = 5\,\Omega$

Hence current is

$I = \frac{\mathcal{E}}{R} = \frac{0.6}{5} = 0.12\,\text{A}$

3. Magnetic force opposing the motion

The magnetic force on the rod is

$F = BIl$

Substituting values,

$F = 0.15 \times 0.12 \times 1 = 0.018\,\text{N}$

This is the retarding force, so the external force needed to move the rod with constant speed must be equal in magnitude:

$F_{\text{ext}} = 0.018\,\text{N}$

4. Expressing in the asked form

We need the answer in units of $10^{-3}\,\text{N}$:

$0.018\,\text{N} = 18 \times 10^{-3}\,\text{N}$

So the required integer is

$\boxed{18}$