Answer: \(1537024\).

First expand the second factor:

\((1-4x^2)^2=1-8x^2+16x^4\).

Now consider

\(\left(2+\frac{3}{x^2}\right)^{10}\).

The general term in this expansion is

\(\binom{10}{r}2^{10-r}\left(\frac{3}{x^2}\right)^r=\binom{10}{r}2^{10-r}3^r x^{-2r}\).

To get a constant term after multiplying by \(1-8x^2+16x^4\), we only need the terms in \(x^0\), \(x^{-2}\) and \(x^{-4}\).

These correspond to \(r=0,1,2\):

\(r=0:\quad 2^{10}=1024\),

\(r=1:\quad \binom{10}{1}2^9\cdot3x^{-2}=15360x^{-2}\),

\(r=2:\quad \binom{10}{2}2^8\cdot3^2x^{-4}=103680x^{-4}\).

The constant term is obtained from

\(1024(1)+(15360x^{-2})(-8x^2)+(103680x^{-4})(16x^4)\).

Therefore the required term is

\(1024-122880+1658880=1537024\).