Answer
${\bf \approx 2.0\times 10^{-18}}\;\rm m$
Work Step by Step
We know that the de Broglie wavelength is given by
$$\lambda=\dfrac{h}{mv}\tag 1$$
So we need to find the mass of the red blood cell.
Recalling that
$\rho =\dfrac{m}{V}$, and hence, $m=\rho V=\rho AL$
where $L$ here is the thickness of the cell.
$$m=\rho (\pi r^2 )L$$
Plug into (1);
$$\lambda=\dfrac{h}{\pi r^2 \rho L v} $$
Plug the known;
$$\lambda=\dfrac{(6.63\times 10^{-34})}{\pi (3.5\times 10^{-6})^2 (1100)(2\times 10^{-6})(4\times 10^{-3})} $$
$$\lambda=\color{red}{\bf 1.96\times 10^{-18}}\;\rm m$$
