b
Let side of triangle is a and mass is \(m\)

\(MOI\) of plate \(ABC\) about centroid

\(I_{0}=\frac{m}{3}\left(\left(\frac{a}{2 \sqrt{3}}\right)^{2} \times 3\right)=\frac{m a^{2}}{12}\)

triangle \(ADE\) is also an equilateral triangle of side \(a / 2\)

Let moment of inertia of triangular plate \(ADE\) about it's centroid \((G')\) is \(I _{1}\) and \(mass\) is \(m _{1}\)

\(m _{1}=\frac{ m }{\frac{\sqrt{3} a ^{2}}{4}} \times \frac{\sqrt{3}}{4}\left(\frac{ a }{2}\right)^{2}=\frac{ m }{4}\)

\(I _{1}=\frac{ m _{1}}{12}\left(\frac{ a }{2}\right)^{2}=\frac{ m }{4 \times 12} \frac{ a ^{2}}{4}=\frac{ ma ^{2}}{192}\)

distance \(G G^{\prime}=\frac{a}{\sqrt{3}}-\frac{a}{2 \sqrt{3}}=\frac{a}{2 \sqrt{3}}\)

so \(MOI\) of part \(ADE\) about centroid \(G\) is

\(I _{2}= I _{1}+ m _{1}\left(\frac{ a }{2 \sqrt{3}}\right)^{2}=\frac{ ma ^{2}}{192}+\frac{ m }{4} \cdot \frac{ a ^{2}}{12}\)

\(=\frac{5 ma ^{2}}{192}\)

now \(MOI\) of remaining part

\(=\frac{m a^{2}}{12}-\frac{5 m a^{2}}{192}=\frac{11 m a^{2}}{12 \times 16}=\frac{11 I_{0}}{16}\)

\(\Rightarrow \quad N=11\)