[tex]\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}[/tex]
[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}[/tex]
[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})[/tex]
[tex]\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}[/tex]
M is molarity here not metre