We know that magnetic field due to finite current carrying wire is

$$B = {{{\mu _0}} \over {4\pi }}{I \over b}(\cos {\theta _1} + \cos {\theta _2})$$ ..... (1)

Given that side of triangle = 4.5 $$\times$$ 10^$$-$$2 m = a; current = 1A since the triangle is equilateral, angle of each side will be 60$$^\circ$$.

Now, $$\tan \theta = {{Perpendicular} \over {Base}} \Rightarrow \tan 60^\circ = {{a/2} \over b}$$

$$ \Rightarrow b = {a \over {2\tan 60^\circ }} = {a \over {2\sqrt 3 }}$$

Using equation (1), we get

$$B = {{{\mu _0}} \over {4\pi }}{I \over {a/2\sqrt 3 }}(\cos 30^\circ + \cos 30^\circ ) = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}2\cos 30^\circ $$

$$B = {{{\mu _0}} \over {4\pi }}{{2\sqrt 3 I} \over a}{{2 \times \sqrt 3 } \over 2} = {{{\mu _0}} \over {4\pi }}{{6I} \over a}$$

$$ \Rightarrow B = {{{{10}^{ - 7}} \times 6 \times 1} \over {4.5 \times {{10}^{ - 2}}}} = 1.33 \times {10^{ - 5}} \sim 2 \times {10^{ - 5}}$$ Wb/m²