Step 1: Derive the induced emf from the rate of change of flux, instead of quoting $\mathcal{E} = BLv$ directly.
As the wire of length $L$ slides with speed $v$ perpendicular to the field, in a small time $dt$ it sweeps an extra strip of area $dA = L(v\,dt)$. The extra flux linked is $d\Phi = B\,dA = BLv\,dt$, so by Faraday's law, \[ \mathcal{E} = \frac{d\Phi}{dt} = BLv \]
Step 2: Convert the given speed to SI units.
\[ v = 180\ \text{m/min} = \frac{180}{60} = 3\ \text{m/s}, \qquad L = 20\ \text{cm} = 0.2\ \text{m} \]
Step 3: Substitute into the flux-rule result.
\[ 3 = B \times 0.2 \times 3 = 0.6B \]
Step 4: Solve for $B$.
\[ B = \frac{3}{0.6} = 5\ \text{T} \]
Final Answer:
\[ \boxed{5\ \text{T}} \]