Consider the system as shown below

where, y(t)=x(e^t). The system is

The correct option is linear and non-causal.

Given the block diagram representing a system with input $x\left(t\right)$ and output $y\left(t\right)$:

The input-output relationship of the system is given by:
$y\left(t\right)=x\left(e^t\right)$
To determine the characteristics of the system, we analyze both its linearity and its causality.

1. Linearity Test:
A system is linear if it satisfies the principle of superposition (homogeneity and additivity).
Let $y_1\left(t\right)$ be the output corresponding to an input $x_1\left(t\right)$:
$y_1\left(t\right)=x_1\left(e^t\right)$
Let $y_2\left(t\right)$ be the output corresponding to another input $x_2\left(t\right)$:
$y_2\left(t\right)=x_2\left(e^t\right)$
Now, let us apply a linear combination of the two inputs as the new input, $x_3\left(t\right)=a{x}_1\left(t\right)+b{x}_2\left(t\right)$, where $a$ and $b$ are constants.
The corresponding output $y_3\left(t\right)$ is:
$y_3\left(t\right)=x_3\left(e^t\right)=a{x}_1\left(e^t\right)+b{x}_2\left(e^t\right)$
Substituting the expressions for $y_1\left(t\right)$ and $y_2\left(t\right)$, we get:
$y_3\left(t\right)=a{y}_1\left(t\right)+b{y}_2\left(t\right)$
Since the system satisfies the superposition principle, it is linear.

2. Causality Test:
A system is causal if the output at any time $t$ depends only on present and/or past values of the input. If the output depends on any future values of the input, the system is non-causal.
Let us evaluate the output at a few specific time instants:
For $t=0$:
$y\left(0\right)=x\left(e^0\right)=x\left(1\right)$
Here, the output at $t=0$ depends on the input at the future time $t=1$.
For $t=2$:
$y\left(2\right)=x\left(e^2\right)\approx x\left(7.39\right)$
Here, the output at $t=2$ depends on the input at the future time $t\approx 7.39$.
Since the output depends on future values of the input (because $e^t>t$ for all real $t$), the system is non-causal.

Conclusion:
The system is both linear and non-causal.