Given polynomial: $p\left(x\right)=\left(5x²-6x+k\right)$ 

If $\alpha \text{ and }\beta$ are the zeroes of $\left(a{x}^2+bx+c\right),$ then
$\alpha +\beta =-\frac{b}{a} \text{and }\alpha \beta =\frac{c}{a}$ 

$\alpha +\beta =\frac{6}{5} \mathrm{and} \alpha \beta =\frac{k}{5}$ -----(1)
Also, given that,

$\alpha -\beta =\frac{1}{5}$ 

Squaring both sides, we get, 

${\left(\alpha -\beta \right)}^2={\left(\frac{1}{5}\right)}^2$

$\Rightarrow \left({\alpha }^2+{\beta }^2-2\alpha \beta \right)=\frac{1}{25}$

$\Rightarrow \left({\alpha }^2+{\beta }^2+2\alpha \beta \right)-2\alpha \beta -2\alpha \beta =\frac{1}{25}$  

$\Rightarrow {\left(\alpha +\beta \right)}^2-4\alpha \beta =\frac{1}{25}$

$\Rightarrow {\left(\frac{6}{5}\right)}^2-4\left(\frac{k}{5}\right)=\frac{1}{25}$ 

$\Rightarrow -\frac{4k}{5}=\frac{1}{25}-\frac{36}{25}$

$\Rightarrow -\frac{4k}{5}=\frac{1-36}{25}$  

$\Rightarrow -\frac{4k}{5}=\frac{-35}{25}$

$\Rightarrow -100k=-175$ 

$\Rightarrow k=\frac{175}{100}$

$\Rightarrow k=\frac{7}{4}$ 
Hence, the value of $k$ is $\frac{7}{4}$.