快速浏览源代码（在线提供）可以确认您所遇到的问题：

/* @deprecated Use the type-safer {@link #getParcelable(String, Class)} starting from Android

* {@link Build.VERSION_CODES#TIRAMISU}.
* /

@Deprecated
@Nullable
public <T extends Parcelable> T getParcelable(@Nullable String key) {
  //Implementation here
}

正如您所看到的，注解建议您使用另一种（类型更安全的）方法：

/**
 * Returns the value associated with the given key or {@code null} if:
 * <ul>
 *     <li>No mapping of the desired type exists for the given key.
 *     <li>A {@code null} value is explicitly associated with the key.
 *     <li>The object is not of type {@code clazz}.
 * </ul>
 *
 * <p><b>Note: </b> if the expected value is not a class provided by the Android platform,
 * you must call {@link #setClassLoader(ClassLoader)} with the proper {@link ClassLoader} first.
 * Otherwise, this method might throw an exception or return {@code null}.
 *
 * @param key a String, or {@code null}
 * @param clazz The type of the object expected
 * @return a Parcelable value, or {@code null}
 */
@SuppressWarnings("unchecked")
@Nullable
public <T> T getParcelable(@Nullable String key, @NonNull Class<T> clazz) {
    // The reason for not using <T extends Parcelable> is because the caller could provide a
    // super class to restrict the children that doesn't implement Parcelable itself while the
    // children do, more details at b/210800751 (same reasoning applies here).
    return get(key, clazz);
}

新方法甚至有一个注解，说明为什么要使用这个方法。
请特别注意：

• @param clazz需要的对象类型

因此，为了回答您的问题，您应该执行以下操作来获取对象：

User MyUser = new User("", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "");
MyUser = bundlee.GetParcelable("MyUser", Java.Lang.Class.FromType(typeof(User))) as User;

或者：

MyUser = bundlee.GetParcelable("MyUser", Java.Lang.Class.FromType(MyUser.GetType())) as User;