$$\tan 30 = {x \over d}$$

$$d = {x \over {\tan 30}}$$

$$d = {{5 \times {{10}^{ - 2}}} \over {{1 \over {\sqrt 3 }}}}$$

$$d = 5\sqrt 3 \times {10^{ - 2}}$$

For one part of the wire with N turns,

$$B = {{{\mu _0}IN} \over {4\pi d}}(\sin {\theta _1} + \sin {\theta _2})$$

For 6 identical parts of the wire,

$${B_{net}} = 6B$$

$$ = {{6{\mu _0}IN} \over {4\pi d}}(\sin 30 + \sin 30)$$

$$ = {{{\mu _0}I} \over \pi }\left( {{{6 \times 50} \over {4 \times 5\sqrt 3 \times {{10}^{ - 2}}}}} \right)\left( {2 \times {1 \over 2}} \right)$$

$$ = 500\sqrt 3 \left( {{{{\mu _0}I} \over \pi }} \right)$$