= Solution
First interpret $\lambda$ as the material wavelength $2\pi/k$. Since $\omega=kv_p(k)$,
$$
v_g=v_p+k\frac{dv_p}{dk}=v_p-\lambda\frac{dv_p}{d\lambda}.
$$
For $v_p=c/(a-b\lambda)$ this gives
$$
\boxed{v_g=\frac{c(a-2b\lambda)}{(a-b\lambda)^2},\qquad \frac{v_g}{v_p}=\frac{a-2b\lambda}{a-b\lambda}.}
$$
With the supplied numbers, $b\lambda=0.015$ and the ratio is $1.470/1.485=0.989899\simeq0.99$.
There is a common optical convention to distinguish. If the dispersion law is instead parameterized by \i[vacuum] wavelength $\lambda_0=2\pi c/\omega$, its refractive index is $n(\lambda_0)=a-b\lambda_0$. The <group refractive index> is then $n_g=n-\lambda_0dn/d\lambda_0=a$, giving $v_g=c/a$ and $v_g/v_p=(a-b\lambda_0)/a=0.990000$. Both conventions yield the requested approximate ratio, but their exact algebra differs. This is the distinction between <group velocity in material and vacuum wavelength coordinates>.
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