The area under the density is a triangle of base 6 and height \(a\). Normalisation gives \[\frac12(6)a=1\quad\Longrightarrow\quad a=\frac13.\]

The line passes through \((0,1/3)\) and \((6,0)\), so \[f(x)=\frac13-\frac{x}{18},\qquad0\le x\le6.\]

Find the second moment by integration: \[\begin{aligned}\mathrm E(X^2)&=\int_0^6x^2\left(\frac13-\frac{x}{18}\right)\,\mathrm dx\\&=\left[\frac{x^3}{9}-\frac{x^4}{72}\right]_0^6\\&=24-18=6.\end{aligned}\]

Use the given mean \(\mathrm E(X)=2\): \[\operatorname{Var}(X)=\mathrm E(X^2)-[\mathrm E(X)]^2=6-2^2=\boxed{2}.\]