I am not sure if I understood your question correctly, but for each random variable $X$ with positive variance, there of course is such a constant $c(X)$ depending on $X$. 

However, there is no uniform constant that works for all random variables with unit variance, as the following counterexample shows. 

For $n \in \mathbb{N}$ consider a random variable $X_n$ with $\mathbb{P}[X_n = n] = \mathbb{P}[X_n = -n] = \frac{1}{2n^2}$ and $\mathbb{P}[X_n = 0] = 1 - \frac{1}{n^2}$. 
Then we have $\mathbb{E}[X_n]=0$ and $\mathbb{E}[X_n^2]=1$ for all $n$, but also $\mathbb{E}[\lvert X_n \rvert] = \frac{1}{n}$. Therefore there cannot be a uniform lower bound $c$ without any additional assumptions.