Given the velocity vector of a particle $v = (2t \hat{i}+3 t^{2} \hat{j}) \, \text{ms}^{-1}$, the acceleration $a$ is the derivative of the velocity vector with respect to time. So, we have:

$a = \frac{dv}{dt} = (2 \hat{i} + 6t \hat{j}) \, \text{ms}^{-2}$.

At $t=1 \, \text{s}$, the acceleration $a$ is $(2 \hat{i} + 6 \hat{j}) \, \text{ms}^{-2}$.

According to Newton's second law, the force $F$ is equal to the mass $m$ times acceleration $a$. The mass $m$ is given as $500 \, \text{g}$, or equivalently, $0.5 \, \text{kg}$.

Therefore, the force $F$ on the particle at $t=1 \, \text{s}$ is:

$F = m \cdot a = 0.5 \cdot (2 \hat{i} + 6 \hat{j}) = (1 \hat{i} + 3 \hat{j}) \, \text{N}$.

So, the force acting on the particle at $t=1 \, \text{s}$ is $(\hat{i} + x \hat{j}) \, \text{N}$, where $x=3$.

Therefore, the answer is $x=3$.