Finding required drug concentration given Kdiss
Chemical reaction is

$$\ce{D1 + E &lt;=&gt; ED1}$$

Initial concentrations: $[\ce{D1}]_\textrm{tot}; \quad [\ce{E}]_\textrm{tot}; \quad 0$.

If $X$ percent of $E$ "reacts", then stoichiometry tells us that:

Concentrations at equilibrium: $[\ce{D1}]_\textrm{tot} (1- X) ;\quad [\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X;  \quad [\ce{D1}]_\textrm{tot} X$.

The equilibrium expression is

$$K_\textrm{diss} = \frac{ [E][D_1]}{[ED_1]}$$

Therefore

$$K_\textrm{diss} = \frac{ [\ce{D1}]_\textrm{tot} (1- X) ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)}{[\ce{D1}]_\textrm{tot} X}$$

Which we can simplify to

$$K_\textrm{diss} = \frac{(1- X)}{X} ([\ce{E}]_\textrm{tot} - [\ce{D1}]_\textrm{tot} X)$$

So you can solve this for $X = 0.90$, $[\ce{E}]_\textrm{tot} = 2.00 \ \mu M$ and $K = 1.00 \ nM$.