The correct answer is 1 (representing the maximum step size h for which the Euler's method is stable).

Step-by-Step Derivation and Explanation:

1. Identify the given differential equation:
We are given the first-order ordinary differential equation (ODE) shown in the image:
$\frac{dy}{dt}+2y=0$
Rewriting it to express the derivative explicitly, we get:
$\frac{dy}{dt}=-2y=f(t,y)$

2. Apply Euler's method:
The standard formula for Euler's forward method with step size h is:
$y_{n+1}=y_n+hf(t_n,y_n)$
Substituting $f(t_n,y_n)=-2y_n$ into the equation:
$y_{n+1}=y_n+h(-2y_n)$
Factoring out $y_n$, we obtain:
$y_{n+1}=y_n(1-2h)$

3. Determine the stability criterion:
For the numerical solution to remain stable and not grow unboundedly as n increases, the magnitude of the solution at step n + 1 must not exceed that at step n:
$\left|y_{n+1}\right|\le \left|y_n\right|$
Substituting the relation for $y_{n+1}$ gives:
$\left|y_n\right(1-2h\left)\right|\le \left|y_n\right|$
Assuming $y_n\ne 0$, we divide both sides by $\left|y_n\right|$:
$|1-2h|\le 1$

4. Solve the inequality for step size h:
Unfolding the absolute value inequality yields:
$-1\le 1-2h\le 1$
Subtracting 1 from all parts of the inequality:
$-2\le -2h\le 0$
Multiplying by -1 and reversing the inequality signs:
$2\ge 2h\ge 0$
Dividing by 2:
$1\ge h\ge 0$
Which is equivalently written as:
$0\le h\le 1$

Hence, the maximum step size for stability is 1.