$$\begin{gathered} \mathrm{If}\mathrm{probability}\mathrm{that}a{x}^2+2\sqrt{2}bx+c>0\forall x\in R \\ \mathrm{where}a,b,c\in \{1,2,3,4\} \\ \mathrm{is}\frac{m}{n}\mathrm{where}m\&n\mathrm{are}\mathrm{coprime} \\ \mathrm{then}(m+n)\mathrm{is} \end{gathered}$$

The correct answer is 81.

We are given the quadratic expression:
$a{x}^2+2\sqrt{2}bx+c>0\forall x\in R$
where $a,b,c\in \{1,2,3,4\}$.

For a quadratic expression $a{x}^2+Bx+C>0$ to be positive for all real numbers $x$, the following two conditions must be satisfied:
1. The coefficient of $x^2$ must be positive: $a>0$. Since $a\in \{1,2,3,4\}$, this condition is always satisfied.
2. The discriminant ($D$) of the quadratic expression must be negative: $D<0$.

Let's calculate the discriminant:
$D={\left(2\sqrt{2}b\right)}^2-4ac$
$D=8b^2-4ac$

For $D<0$:
$8b^2-4ac<0$
$2b^2<ac$
or equivalently, $ac>2b^2$.

The elements $a,b,c$ are chosen from the set $\{1,2,3,4\}$. The total number of outcomes for $(a,b,c)$ is:
$\text{Total outcomes}=4\times 4\times 4=64$

Now, let's analyze the cases based on the value of $b$ to find the number of favorable outcomes:

Case 1: $b=1$
Here, $2b^2=2{\left(1\right)}^2=2$.
We require $ac>2$.
The pairs $(a,c)$ that do not satisfy this condition (i.e., $ac\le 2$) are:
- $(1,1)$ since $1\times 1=1$
- $(1,2)$ since $1\times 2=2$
- $(2,1)$ since $2\times 1=2$
There are 3 such pairs. Since there are $4\times 4=16$ possible pairs of $(a,c)$ in total, the number of favorable pairs for $b=1$ is:
$16-3=13$

Case 2: $b=2$
Here, $2b^2=2{\left(2\right)}^2=8$.
We require $ac>8$.
Let's find the pairs $(a,c)$ that satisfy $ac>8$:
- If $a=3$, then $c$ can be 3 or 4 (giving products 9 and 12) ⇒ 2 pairs: $(3,3),(3,4)$
- If $a=4$, then $c$ can be 3 or 4 (giving products 12 and 16) ⇒ 2 pairs: $(4,3),(4,4)$
Thus, the number of favorable pairs for $b=2$ is 4.

Case 3: $b=3$
Here, $2b^2=2{\left(3\right)}^2=18$.
We require $ac>18$.
Since the maximum possible value of $ac$ is $4\times 4=16$, there are no pairs that satisfy this condition. Thus, the number of favorable pairs is 0.

Case 4: $b=4$
Here, $2b^2=2{\left(4\right)}^2=32$.
We require $ac>32$.
Since the maximum value of $ac$ is 16, the number of favorable pairs is 0.

Now, let's sum up the favorable outcomes:
$\text{Total Favorable Outcomes}=13+4+0+0=17$

The probability is given by:
$P=\frac{m}{n}=\frac{17}{64}$

Since 17 is a prime number and 64 is not divisible by 17, $m=17$ and $n=64$ are coprime.

Thus, we compute the required value:
$m+n=17+64=81$