Answer: \(a=\dfrac15\), \(b=243\), \(c=\dfrac{54}{5}\).

Use the binomial expansion of

\((3-ax)^5.\)

The constant term is

\(3^5=243,\)

so

\(b=243.\)

The coefficient of \(x\) is

\(5(3^4)(-a)=-405a.\)

This is equal to \(-81\), so

\(-405a=-81.\)

Hence

\(a=\frac15.\)

The coefficient of \(x^2\) is

\({5\choose2}3^3a^2.\)

So

\(c=10\cdot27\cdot\left(\frac15\right)^2 =270\cdot\frac1{25} =\frac{54}{5}.\)