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\(B = \frac{F}{q \cdot v \cdot \sin \theta}\)

\(\theta = 45^\circ \implies \sin 45^\circ = \frac{\sqrt{2}}{2}\)

Substituting the values:

\(B = \frac{6}{(5 \times 10^{-6})(3 \times 10^4) \left(\frac{\sqrt{2}}{2}\right)}\)

Simplifying:

\(B = \frac{6}{(15 \times 10^{-2}) \left(\frac{\sqrt{2}}{2}\right)} \approx \frac{6}{0.1061} \approx 56.5 \, \text{T}\)