What is the minimum pressure for existence of a liquid solution?
The only things I can figure for $a$ and $b$ in $p=\dfrac{a}{27b^2}$ are the coefficients from the van der Waals equation for nonideal gases:

$$\left(p+\dfrac{n^2a}{V^2}\right)\left(V-nb\right)=nRT$$

These constants have appropriate units to give you pressure: $a\ \left( \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}\right)$ and $b \ \left(\dfrac{\text{L}}{\text{mol}} \right)$. 

$$\dfrac{a}{27b^2}\implies\dfrac{\dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left(\dfrac{\text{L}}{\text{mol}}\right)^2}=\dfrac{\text{atm}}{27}\implies p$$

However, these constants are empirically derived, and thus different for every substance, and so the critical pressure would be different for every substance.

For example, for argon $a=1.355 \ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}$ and $b=0.0320\ \dfrac{\text{L}}{\text{mol}}$, so $p$ for argon is 

$$p=\dfrac{1.355\ \dfrac{\text{L}^2 \ \text{atm}}{\text{mol}^2}}{27\left( 0.0320\ \dfrac{\text{L}}{\text{mol}} \right)^2}=49.01\ \text{atm}$$

Were you told which substance you should be using? For all substances, $b$ appears to be less than $1$, so we need to find the minimum $a$ and maximum $b$ combination to get the minimum $p$. However, by my calculation, all values of $p$ will be greater than $1$. The value for water is $220.5\text{ atm}$ (which does not make sense given that a vast number of liquid solutions are stable at normal atmosphere pressure).